Image of Convex Set under Linear Transformation is Convex

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ and $Y$ be vector spaces over $\Bbb F$.

Let $C \subseteq X$ be convex.

Let $T : X \to Y$ be a linear transformation.

Then $\map T C \subseteq Y$ is convex.

Proof

Let $y_1, y_2 \in \map T C$ and $\lambda \in \closedint 0 1$.

Then, there exists $x_1, x_2 \in C$ such that:

$y_1 = T x_1$

and:

$y_2 = T x_2$

Then, we have:

$\lambda y_1 + \paren {1 - \lambda} y_2 = \lambda T x_1 + \paren {1 - \lambda} T x_2 = \map T {\lambda x_1 + \paren {1 - \lambda} x_2}$

Since $C$ is convex, we have:

$\lambda x_1 + \paren {1 - \lambda} x_2 \in C$

So, we have:

$\map T {\lambda x_1 + \paren {1 - \lambda} x_2} \in \map T C$

That is:

$\lambda y_1 + \paren {1 - \lambda} y_2 \in \map T C$

So $\map T C$ is convex.

$\blacksquare$